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3 years ago

Use the slope of the given line to construct a parallel line

Mathematics

2 answers:

Troyanec [42]3 years ago

6 0

Answer:

y = - $\frac{4}{3}$ x - 7

Step-by-step explanation:

Parallel lines have equal slopes.

Calculate the slope m of the given line using the slope formula

m = $\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

with (x₁, y₁ ) = (0, 2) and (x₂, y₂ ) = (3, - 2) ← 2 points on the line

m = $\frac{-2-2}{3-0}$ = - $\frac{4}{3}$ , thus

y = - $\frac{4}{3}$ x + c ← is the partial equation

To find c substitute P(- 3, - 3) into the partial equation

- 3 = 4 + c ⇒ c = - 3 - 4 = - 7

y = - $\frac{4}{3}$ x - 7 ← equation of parallel line

tatuchka [14]3 years ago

3 0

Answer:

put the point on position -2, -4

Step-by-step explanation: i just finished this question. and this was the right option

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3 years ago

A website sells a dress at £40 and shoes at £32. The website has an offer: Buy a dress and shoes together and get 1/2 off the pr

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Answer:

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Step-by-step explanation:

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2 years ago

A square garden has 49 square feet. So the perimeter of the garden must be----feet.

lianna [129]

The side length of a square is the square root of the area:

Side length = √49 = 7 feet.

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2 years ago

Read 2 more answers

[(e^x -e^-x)/(e^x + e^-x)] = 1/3

mr_godi [17]

Answer:

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Step-by-step explanation:

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3 years ago

Find the slope of the tangent line to the given polar curve at the point specified by the value of θ

MariettaO [177]

The given curve has equation

r(θ) = 9 + 8 cos(θ)

and its derivative is

dr/dθ = -8 sin(θ)

When θ = π/3, we have r (π/3) = 13, and dr/dθ (π/3) = -4√3.

Differentiate these with respect to θ :

dy/dθ = dr/dθ sin(θ) + r(θ) cos(θ)

dx/dθ = dr/dθ cos(θ) - r(θ) sin(θ)

In polar coordinates, we have

y(θ) = r(θ) sin(θ)

x(θ) = r(θ) cos(θ)

and when θ = π/3, we have y (π/3) = 13√3/2 and x (π/3) = 13/2.

The slope of the tangent line to the curve is dy/dx. By the chain rule,

dy/dx = dy/dθ • dθ/dx = (dy/dθ) / (dx/dθ)

dy/dx = (dr/dθ sin(θ) + r(θ) cos(θ)) / (dr/dθ cos(θ) - r(θ) sin(θ))

When θ = π/3, the slope is

dy/dx = (-4√3 sin(π/3) + 13 cos(π/3)) / (-4√3 cos(π/3) - 13 sin(π/3))

dy/dx = (-4√3 (√3/2) + 13 (1/2)) / (-4√3 (1/2) - 13 (√3/2))

dy/dx = - 1/(17√3)

So, the tangent line has slope -1/(17√3) and passes through (13/2, 13√3/2). Using the point-slope formula, its equation is

y - 13√3/2 = -1/(17√3) (x - 13/2)

y = -(x - 338)/(17√3)

4 0

2 years ago